Question

$\frac{x}{\cos \alpha }=\frac{y}{\cos (\alpha -\frac{2\pi }{3})}=\frac{z}{\cos (\alpha +\frac{2\pi }{3})}$
then x + y + z is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{1}{4}\cos 3\alpha$

Answer 1

The correct option is B $0$

Let $\frac{x}{\cos \alpha }=\frac{y}{\cos (\alpha -\frac{2\pi }{3})}=\frac{y}{\cos (\alpha +\frac{2\pi }{3})}=k$

Now, $x=k\cos \alpha$
$y=k\cos (\alpha -\frac{2\pi }{3})$
$z=k\cos (\alpha +\frac{2\pi }{3})$

Now, $x+y+z=k\cos \alpha +k\cos (\alpha -\frac{2\pi }{3})+k\cos (\alpha +\frac{2\pi }{3})$
$x+y+z=k\cos \alpha +k\cos \alpha \cos \frac{2\pi }{3}+k\sin \alpha \sin \frac{2\pi }{3}+k\cos \alpha \cos \frac{2\pi }{3}-k\sin \alpha \sin \frac{2\pi }{3}$
$x+y+z=k\cos \alpha +2k\cos \alpha \cos \frac{2\pi }{3}$
$x+y+z=k\cos \alpha +2k\cos \alpha \left(\frac{-1}{2}\right)$
Hence, $x+y+z=0$